Differentiable unbounded function $f: (a, b)\to \mathbb R$ must have unbounded derivatives For a function $f: (a, b) \to \mathbb{R}$ that is unbounded and differentiable, how can you show, using the Mean Value Theorem, that its derivative is also an unbounded function?
As far as I have gotten is basically just the definition of unboundedness being when the function is not bounded.
Bounded meaning that $\exists M \in \mathbb{R}:  |f(x)| \leq M, \forall x \in X$, where $X$ is the set the function is defined on.
And the Mean Value Theorem being $f'(c) = \frac{f(b) - f(a)}{b-a}$ for some interval $(a,b)$.
But I am unsure how this applies to the whole function, $f(x)$, not just at the point $c$, and how this can be used to show that the derivative is unbounded.
 A: Note that it is crucial the domain of the function be a bounded open interval, because if the domain is $\Bbb{R}$ then the claim is false ($f(x)=x$ is unbounded on $\Bbb{R}$ yet its derivative is constant hence trivially bounded).
One way of phrasing the argument is contrapositively, and I find that simpler. So, we suppose $f'$ is bounded on $(a,b)$, say by $B$, and we want to show $f$ itself is bounded. So, fix an $x_0\in(a,b)$ (for example $\frac{a+b}{2}$... it doesn't really matter, just pick a point and keep it fixed for the rest of the discussion), and let $x\in (a,b)$ be arbitrary. Then,
\begin{align}
|f(x)-f(x_0)|\leq B|x-x_0|.
\end{align}
(if $x=x_0$ this inequality is trivially true, if $x<x_0$ use the mean-value theorem on the interval $(x,x_0)$ while if $x_0<x$ then use the mean-value theorem on $(x_0,x)$). In particular, by the reverse triangle inequality,
\begin{align}
|f(x)|&\leq |f(x_0)|+B|x-x_0|\\
&\leq |f(x_0)|+B(b-a).
\end{align}
Since $x\in (a,b)$ is arbitrary, we have that $f$ is bounded by $M:=|f(x_0)|+B(b-a)$.
The crux of this argument is that if $f'$ is bounded, then $f$ satisfies a Lipschitz condition, and therefore is bounded on every bounded interval.
A: Hints:
Since f is unbounded there is point $x_0$ such that $ |f(x_0)| > M$ for any $M >0.$ Now choose another point $x,$ which is different from $x_0$ (WLOG, we can make it smaller than $x_0$) such that $f(x)= k$ for some finite $k.$ Then apply MVT on the interval $ (x, x_0).$
A: Fix $M\in(0,\infty)$. Since $f$ is unbounded, then there exists an $x_1\in (a,b)$ such that $|f(x_1)|>M$. There also exists an $x_2\in (a,b)$ such that $|f(x_2)|>|f(x_1)|+M(b-a)$. By the mean value theorem, there exists a $c$ between $x_1$ and $x_2$ such that $$f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$
Now we have
$$\begin{align*}|f'(c)|
&=\left|\frac{f(x_2)-f(x_1)}{x_2-x_1}\right|\\
&\geq \frac{|f(x_2)|-|f(x_1)|}{|x_2-x_1|}\\
&> M\frac{(b-a)}{|x_2-x_1|}\\
&>M \end{align*}$$
since we have $$b-a>|x_2-x_1|>0\Longrightarrow \frac{b-a}{|x_2-x_1|}>1 $$
A: Let $c=(a+b)/2$ and let $d=(b-a)/2.$
For any $r>0,$ take some  $x\in (a,b)$ with $|f(x)|>|f(c)|+d\cdot r.$ Then $x\ne c$ because $|f(x)|\ne|f(c)|.$ So we have $$\frac {|f(x)-f(c)|}{|x-c|}\ge \frac {|f(x)|-|f(c)|}{|x-c|}>$$ $$>\frac {d\cdot r}{|x-c|}>$$ $$>\frac {d\cdot r}{d}=r.$$ And there exists $y$ between $x$ and $c$ with $\frac {|f(x)-f(c)|}{|x-c|}=|f'(y)|.$
